Question

Out of 21 tickets marked with numbers from1 to 21, three are drawn at random. The chance that the numbers on them are in A.P. is

• A. $\frac { 9 } { 1330 }$
• B. $\frac { 9 } { 133 }$
• C. $\frac { 10 } { 133 }$
• D. None of these

Answer 1

Answer: (C)

$\frac { 10 } { 133 }$

Answer 2

Total ways = ²¹C₃ = $\frac { 21 \times 20 \times 19 } { 6 }$ = 1330

No. of AP. with common digits No. of way

d = 1, (1, 2, 3), (2, 3, 4) ……(19, 20, 21) ® 19

d = 2, (1, 3, 5), (2, 4, 6) ……. (17, 19, 21) ® 7

d = 3, (1, 4, 7), (2, 5, 8)….... (15, 18, 12) ® 15

d = 4,

d = 5,

d = 6,

d = 7,

d = 8, (1, 9, 17) (2, 10, 18), (3, 11, 19) (4, 13, 21) ® 4

d = 9, (1, 10, 19) (2, 11, 20), (3, 12, 21) ® 3

d = 10, (1, 11, 21) ® 1

Total AP = 1 + 3 + 5 + …….+19 = 100

Prob. = $\frac { 100 } { 1330 }$ = $\frac { 10 } { 133 }$